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I got a bit confused while checking an answer for simple equation: $$x^3 = 1.$$

I got $$x = 1, x = e^{j*\pi/3}, x = e^{-j*\pi/3},$$ while the asnwer in the solution is $$x = 1, x = e^{j*2\pi/3}, x = e^{-j*2\pi/3}.$$

Do I understand correctly that $$x = e^{j*2\pi/3} = e^{-j*\pi/3}$$ $$x = e^{-j*2\pi/3} = e^{j*\pi/3}?$$

Tbh, my skills of trigonometry are a bit rusty already, so I'm not completely sure and wanted to double check. Thanks.

aquila
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  • Try to format your equations using MathJax, it will make your question much more readable, and therefore make it more likely for someone to answer it :) – Zubin Mukerjee Jun 04 '17 at 22:09
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    I got ... x = exp(j pi/3) Then $x^3=e^{3 ,\cdot j \pi/3}=e^{j \pi},$ Does that equal $1,$, as it should? – dxiv Jun 04 '17 at 22:16
  • Try cubing your solutions and seeing if they equal one. Bear in mind that exp(jpi/3) cubed is exp(jpi), which is equal to -1 by Euler... – Hayden Jun 04 '17 at 22:16
  • You should multiply your exponents by $2$. – Friedrich Philipp Jun 04 '17 at 22:22
  • Yes, you're right. Just improved formatting. As for the question itself, I've checked with MATLAB just now - the solutions are ok, but I want to know if they are really equivalent (and equivalent in the way I put in my question, not in another way) to the solutions provided by the manual. So the question is just about trigonometry and trigonometric formulas that should be used for proving/disproving it, not about the equation itself. Thanks. – aquila Jun 04 '17 at 22:23
  • From all the formulas I've seen while searching in the internet, $x=e^{j\theta}$, there $\theta = argtan(b/a)$. Why multiply by 2? – aquila Jun 04 '17 at 22:24
  • @Valeria: $1=e^{2j\pi}$, not $e^{j\pi}$. –  Jun 04 '17 at 22:28
  • @Valeria I've checked with MATLAB just now - the solutions are ok No, they are definitely not OK. Whatever you checked in MATLAB must have been something else. – dxiv Jun 04 '17 at 22:28
  • Well, I've put $a = exp(-j*pi/3)$ and then $a^{3}$. So I got $ans = -1.0000 - 0.0000i$. Same result for all other roots (both from my solution and from the manual). – aquila Jun 04 '17 at 22:32
  • @Valeria I got ans=−1.0000−0.0000i Precisely. But you were solving the equation $x^3=\color{red}{1}$ and $\color{red}{−1.0000}−0.0000i \ne 1,$, which proves in fact that a=exp(−j∗pi/3) is not a solution. – dxiv Jun 04 '17 at 22:33
  • Omg, I somehow missed the minus there completely. You´re right. – aquila Jun 04 '17 at 22:36
  • See https://math.stackexchange.com/questions/192742/how-to-solve-x3-1 – lab bhattacharjee Jun 05 '17 at 01:31

3 Answers3

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$$\log x^3=3\log x=\log 1=0+j2k\pi$$ so that

$$\log x=j\frac{2k\pi}3.$$

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Not quite!

$\large e^{\frac{2πj}{3}}=e^{-\frac{4πj}{3}}$ and $\large e^{-\frac{2πj}{3}}=e^{\frac{4πj}{3}}$.

Remember that $e^{2πj}=1$, so dividing or multiplying it won't change the value of the exponential.

Also, speaking of trigonometry, remember that the same $\theta$ that produces $(\cos(\theta),\sin(\theta))$ also makes $e^{j\theta}=\cos(\theta)+j\sin(\theta)$.

So if you think about that, you could gain some intuition on why the exponential repeats modulo $2π$.

Saketh Malyala
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Your solutions are wrong.

$$(\exp{i\pi/3})^3 = \exp{i \pi} = -1$$

In general though - if you have a solution that works, you can add(or remove) $2 \pi$ to the exponent and get the same answer.

Piotr Benedysiuk
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